Let $c,d$ be natural numbers of same parity (both odd or both even) and $\sigma$ be sum of divisors function. Is it known whether or under what conditions $\sigma (c^{2})$=$\sigma (d^{2})$ ? I am guessing that it can happen but unsure if certain things must hold

  • $\begingroup$ Have you seen this? $\endgroup$ – Don Thousand Dec 24 '18 at 19:19
  • $\begingroup$ Thank you,I don't know if that helps me $\endgroup$ – argamon Dec 24 '18 at 19:21
  • $\begingroup$ @argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$ $\endgroup$ – AgentS Dec 24 '18 at 20:08
  • $\begingroup$ I bet you had tried that, not that interesting.. $\endgroup$ – AgentS Dec 24 '18 at 20:12
  • $\begingroup$ Haha, it's just more useful if I know in general but I greatly appreciate your help $\endgroup$ – argamon Dec 24 '18 at 20:13

It looks to me like $\sigma(627^2)=\sigma(749^2)$.

Feel free to check my work. (Or Python's work!)

And here's an even-even pair: $\sigma(740^2)=\sigma(878^2)$.

| cite | improve this answer | |

As pointed out in the comments, starting from the simple example $$\sigma (4^2)=\sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$\sigma(12^2)=(1+2+2^2+2^3+2^4)\times (1+3+3^2)=31\times 13=403$$ $$\sigma(15^2)=(1+3+9)\times (1+5+25)=13\times 31=403$$

and so on.

  • $\begingroup$ That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even $\endgroup$ – argamon Dec 24 '18 at 19:45
  • $\begingroup$ I suggest doing a search. I also found $\sigma(76^2)=\sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that. $\endgroup$ – lulu Dec 24 '18 at 19:48
  • $\begingroup$ In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair. $\endgroup$ – Ross Millikan Dec 24 '18 at 20:22
  • $\begingroup$ You have $\sigma(4^2)=\sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$ $\endgroup$ – Ross Millikan Dec 24 '18 at 20:44
  • $\begingroup$ @RossMillikan Yes, both my examples are simple consequences of $\sigma(4^2)=\sigma (5^2). I'll edit to point that out, $\endgroup$ – lulu Dec 24 '18 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.